Let $(S,*)$ be a finite semigroup with identity. Prove that $S$ is a group iff $S$ has only one element $x$ such that $x^2=x$.

Question

Attempt: Does this approach true?

$(\Rightarrow)$ Let $S$ be a group. Denote $e_S$ be an identity element in $S$. Clearly, $e_S$ is idempotent since $e_S^2 = e_S$. Suppose that there is another idempotent in $S$, say $x$. Then, $x^2 = x$. So, \begin{equation*} x^2=x \Leftrightarrow x^{-1}x^2 = x^{-1}x \Leftrightarrow x = e_S. \end{equation*} Hence, $x=e_S$. Thus, the only idempotent in $S$ is $e_S$.

$(\Leftarrow)$ First, to show for any $x \in S$, there exist $n \in \Bbb N$ such that $x^n$ is idempotent. I found this proof in the ProofWiki. Next, since the only idempotent in $S$ is $e_S$, then $x^n$ must equal to $e_S$. I have proved this already. But, I am still confused. If $n=2$, then $a^2 = e_S$ i.e. the inverse of $a$ is itself. But, what about $n=3$? $a^3=e_S$, does it implies that the inverse of $a$ is $a^2$?

@user795084 ahh sorry I missed the fact that $S$ is a group here — delusional.existence, Jan 18 '21 at 22:50
@AnikBhowmick Ohhh. I make a small mistakes. Yes, you're true. — lap lapan, Jan 18 '21 at 22:58
Then definitely we can say that $a^2$ is the inverse of $a$. Afterall that's the definition of inverse. The approach is perfect. — delusional.existence, Jan 18 '21 at 23:14

score 4 · Accepted Answer · answered Jan 18 '21 at 22:56

4

Let $x\in S$. Since $S$ is finite, there exist $a$, $d> 0$ such that $$x^a= x^{a+d}$$

Now you can show easily by induction that $$x^b = x^{b+ m d}$$

whenever $b\ge a$, and $m\ge 0$ integer. Let us choose $b\ge a$, and $m d$ such that they are both equal. ( for instance, take $b = \operatorname{lcm}(a,d)$). Then we get $$x^{b} = x^{b+b}$$ or $x^b= (x^b)^2$. Therefore, $x^b$ is idempotent, so it must be $e$. We conclude $x^{b-1} = x^{-1}$.

answered Jan 18 '21 at 22:56

orangeskid

56,630

This is guarantee that every $x$ in $S$ has an inverse, right? – lap lapan Jan 18 '21 at 22:59
@user795084: correct! – orangeskid Jan 18 '21 at 23:00
Where does the $m$ come from, after the induction? – Jan 18 '21 at 23:51
@Chris Custer: Oh, first you do induction on $b$. Then induction on $m$. You need both $b$ and $m$ in order to make $b$ and $m d$ equals. Hope that makes it clearer. – orangeskid Jan 19 '21 at 00:06
Oh. Should have seen that. Pretty slick. – Jan 19 '21 at 00:08

Let $(S,*)$ be a finite semigroup with identity. Prove that $S$ is a group iff $S$ has only one element $x$ such that $x^2=x$.

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